Solve x²+2x-8: Finding Values Where Function is Negative

To solve for when $f(x) = x^2 + 2x - 8$ is less than zero, we follow these steps:

• Step 1: Find the roots of the quadratic equation.

The equation is $x^2 + 2x - 8 = 0$. We apply the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a = 1$, $b = 2$, $c = -8$.

Substitute these values into the formula:

$x = \frac{-2 \pm \sqrt{2^2 - 4 \times 1 \times (-8)}}{2 \times 1}$

$x = \frac{-2 \pm \sqrt{4 + 32}}{2}$

$x = \frac{-2 \pm \sqrt{36}}{2}$

$x = \frac{-2 \pm 6}{2}$

This gives two roots: $x_1 = \frac{4}{2} = 2$ and $x_2 = \frac{-8}{2} = -4$.

• Step 2: Identify the intervals created by these roots.

The roots divide the number line into intervals: $(-\infty, -4)$, $(-4, 2)$, and $(2, \infty)$.

• Step 3: Test within each interval to determine where $f(x) < 0$.

Choose test points in each interval:

For $(-\infty, -4)$, test $x = -5$. $f(-5) = 25 - 10 - 8 = 7$. (Positive)

For $(-4, 2)$, test $x = 0$. $f(0) = 0 + 0 - 8 = -8$. (Negative)

For $(2, \infty)$, test $x = 3$. $f(3) = 9 + 6 - 8 = 7$. (Positive)

Thus, $f(x) < 0$ in the interval $(-4, 2)$.

Therefore, the correct interval for which the function $f(x) < 0$ is

$-4 < x < 2$.